Right Truncation With Left Censoring

Why Truncation and Censoring Must Be Handled Together

Suppose a hypothetical insurer analyzes losses only when the loss is no greater than a fixed reporting threshold. This is right truncation: a loss above the threshold does not enter the observed sample at all. Among the selected losses, some records may be reported only as “at most \(c\),” rather than with an exact amount. This is left censoring. The observed censored record is not a loss equal to \(c\); it tells us that the loss lies at or below \(c\).

These two features affect different parts of the likelihood. Truncation changes the distribution of every sampled record because selection requires the loss to be below the upper threshold. Left censoring changes what is known about a particular record: its contribution is a conditional probability, not a density at the censoring point. The treatment here is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let \(X\) be a continuous loss with cumulative distribution function \(F_X(x;\theta)\) and density \(f_X(x;\theta)\), where \(\theta\) is an unknown parameter or parameter vector. A sample subject to right truncation at \(T\) includes only losses satisfying \(X\le T\). A left-censored record at \(c\) reports the event \(X\le c\), rather than the exact value of \(X\). Assume \(c<T\) for a nontrivial left-censored record.

The likelihood below is conditional on a record having entered the sample. It assumes losses are independent and identically distributed before selection, the upper threshold is fixed, and the censoring rule is noninformative: the record is reported as censored because it falls below the fixed point \(c\), not because of some further loss-dependent selection mechanism. If the count of losses that failed to enter the sample is also observed, that information requires a different likelihood treatment.

Conditional Contributions

For each sampled record, the model must be conditioned on the event that made it observable, \(X\le T\). An exact observed loss \(x\) contributes a conditional density. A record known only to satisfy \(X\le c\) contributes a conditional probability.

Theorem: Under a continuous loss model and the assumptions above, the conditional likelihood contribution for an exact sampled loss \(x\le T\) is \(f_X(x;\theta)/F_X(T;\theta)\). The contribution for a sampled record known only to satisfy \(X\le c<T\) is \(F_X(c;\theta)/F_X(T;\theta)\).

The numerator for an exact observation is a density, while the numerator for a censored observation is a probability. Both have the same truncation normalizer because both records are known to have entered the sample. In particular, the censored record is not assigned \(f_X(c;\theta)\): the observed information is an event, not an exact loss amount.

f_{X\mid X\le T}(x;\theta) = \frac{f_X(x;\theta)}{F_X(T;\theta)} \quad (x\le T), \qquad P(X\le c\mid X\le T) = \frac{F_X(c;\theta)}{F_X(T;\theta)} \quad (c<T).

To justify the exact-observation formula, take a small interval of width \(h\) around an observed value \(x\), within the selected range. The probability of landing in that interval after selection is approximately \(f_X(x;\theta)h/F_X(T;\theta)\). Dividing by \(h\) and taking the limit gives the conditional density. For the censored record, \(X\le c\) is a subset of \(X\le T\), so the definition of conditional probability gives \(P(X\le c\mid X\le T)=P(X\le c)/P(X\le T)\). This yields the CDF ratio directly.

\lim_{h\to 0} \frac{P(x\le X\le x+h\mid X\le T)}{h} = \lim_{h\to 0} \frac{F_X(x+h;\theta)-F_X(x;\theta)} {hF_X(T;\theta)} = \frac{f_X(x;\theta)}{F_X(T;\theta)}.

Suppose a sample contains \(m\) left-censored records, all at the same \(c<T\), and \(n\) exact records \(x_1,\ldots,x_n\). Conditional independence gives the likelihood

L(\theta) = \left(\frac{F_X(c;\theta)}{F_X(T;\theta)}\right)^m \prod_{j=1}^{n} \frac{f_X(x_j;\theta)}{F_X(T;\theta)} = \frac{F_X(c;\theta)^m\prod_{j=1}^{n}f_X(x_j;\theta)} {F_X(T;\theta)^{m+n}}.

The denominator appears once for every sampled record, including every censored record. This is not an optional adjustment: it changes the likelihood as a function of \(\theta\). If a censoring point is at or above \(T\), then every selected loss is already at or below that point, and the censored event has conditional probability one. If \(c\) is below the support of the loss, then \(F_X(c;\theta)=0\); such a censored record has zero likelihood under that model.

Worked Likelihoods

Worked Example: Exponential Losses With Two Censored Records

A hypothetical portfolio is modeled with \(X\sim\operatorname{Exponential}(\lambda)\), where the rate is \(\lambda=0.5\) per thousand dollars. Only losses at or below \(T=4\) thousand dollars are sampled. Among the sampled records, two are known only to be at or below \(c=2\) thousand dollars; two others are observed exactly as \(x_1=2.5\) and \(x_2=3\) thousand dollars. Calculate the likelihood at the specified rate. No interest or mortality assumption is relevant to this severity likelihood.

Method. For an exponential loss, \(F_X(x)=1-e^{-\lambda x}\) and \(f_X(x)=\lambda e^{-\lambda x}\), for \(x\ge0\). The selection probability is

F_X(4)=1-e^{-0.5(4)}=1-e^{-2}=0.8646647.

Each censored record contributes

\frac{F_X(2)}{F_X(4)} = \frac{1-e^{-1}}{1-e^{-2}} = \frac{0.6321206}{0.8646647} = 0.7310586.

For the exact record at \(2.5\), the density is \(0.5e^{-1.25}=0.1432524\), so its conditional density contribution is \(0.1432524/0.8646647=0.1657\). For the exact record at \(3\), the density is \(0.5e^{-1.5}=0.1115651\), so its conditional contribution is \(0.1115651/0.8646647=0.1290\).

There are two censored records, so the full likelihood is

L(0.5) = (0.7310586)^2(0.1657)(0.1290) \approx 0.0114.

As a check, the product is approximately \(0.5344\times0.02138=0.01142\), consistent with \(0.0114\) when rounded to four decimal places. The exact-record factors are densities, so the likelihood itself carries density units; it is not a probability of observing those exact values.

Common mistake avoided: using \(F_X(2)\) for each censored record without dividing by \(F_X(4)\). That would use the unselected distribution even though every record in the data has already passed the upper-truncation rule.

Worked Example: Pareto Losses and a Censored Probability

A hypothetical loss model has a Pareto Type I distribution with minimum loss \(\theta=1\) thousand dollars and shape parameter \(\alpha=2\). Losses are sampled only when \(X\le T=4\) thousand dollars. Two sampled records are known only to satisfy \(X\le c=2\) thousand dollars, and one sampled record is observed exactly at \(x=3\) thousand dollars. Calculate the likelihood at these parameter values.

Method. For \(x\ge\theta\), the Pareto CDF and density are \(F_X(x)=1-(\theta/x)^\alpha\) and \(f_X(x)=\alpha\theta^\alpha/x^{\alpha+1}\). Thus

F_X(4)=1-\left(\frac{1}{4}\right)^2=\frac{15}{16}, \qquad F_X(2)=1-\left(\frac{1}{2}\right)^2=\frac{3}{4}.

The contribution from either censored record is the conditional probability

\frac{F_X(2)}{F_X(4)} = \frac{3/4}{15/16} = \frac{4}{5} = 0.8.

At the exact value \(x=3\), the density is \(2/3^3=2/27\). The conditional density contribution is

\frac{f_X(3)}{F_X(4)} = \frac{2/27}{15/16} = \frac{32}{405} \approx 0.0790.

Multiplying the two censored contributions and the exact contribution gives

L = \left(\frac{4}{5}\right)^2\left(\frac{32}{405}\right) = \frac{512}{10125} \approx 0.0506.

The check is \(0.64\times0.0790123\approx0.050568\), which rounds to \(0.0506\) to four decimal places. The censoring numerator is \(F_X(2)\), not \(F_X(4)-F_X(2)\); the latter is the probability of being above 2 and at or below 4 before conditioning, which describes a different event.

Worked Example: Assemble a Mixed Exponential Likelihood

Consider a hypothetical sample from \(X\sim\operatorname{Exponential}(\lambda)\), with \(\lambda=0.2\) per thousand dollars. Records enter only if \(X\le T=5\) thousand dollars. Two selected records are reported as \(X\le c=2\) thousand dollars, while two are observed exactly at \(3\) and \(4\) thousand dollars. Write and evaluate the likelihood at the specified rate.

Method. First calculate the selection probability:

F_X(5)=1-e^{-0.2(5)}=1-e^{-1}=0.6321206.

The probability contribution for either left-censored record is

\frac{F_X(2)}{F_X(5)} = \frac{1-e^{-0.4}}{1-e^{-1}} = \frac{0.3296800}{0.6321206} \approx0.5215.

The exact density contribution at \(3\) is

\frac{0.2e^{-0.6}}{0.6321206} = \frac{0.1097623}{0.6321206} \approx0.1736,

and at \(4\) it is

\frac{0.2e^{-0.8}}{0.6321206} = \frac{0.0898658}{0.6321206} \approx0.1422.

Therefore,

L(0.2) = (0.5215)^2(0.1736)(0.1422) \approx0.0067.

Using the unrounded ratios gives approximately \(0.0067\) to four decimal places, confirming the product. The likelihood has two copies of the censoring probability and two exact-density factors. In particular, the denominator \(F_X(5)\) is present four times in the expanded likelihood because there are four sampled records.

Common mistake avoided: dividing only the exact-density factors by the selection probability. Each censored record is also observed conditional on selection, so each requires its own \(F_X(5)\) denominator.

Interpreting the Contributions and Avoiding Traps

The likelihood is a product of contributions corresponding to what was actually recorded. An exact loss contributes a conditional density; a left-censored record contributes a conditional probability. Keeping this distinction explicit prevents the common error of treating a censoring point as though it were an observed loss amount.

Write the selection event.
For right truncation at \(T\), each included record satisfies \(X\le T\). Identify the corresponding normalizer \(F_X(T;\theta)\).

Classify each record by its observation.
Use a conditional density for an exact value and a conditional probability for the event \(X\le c\) when the value is left-censored.

Multiply the contributions.
Include the truncation normalizer for every record, then substitute the model CDF and density. Check that each censored event is compatible with the selection rule.

A useful exam check is to ask what event the numerator represents. For a left-censored record, the reported information is \(X\le c\), so the numerator is \(F_X(c;\theta)\). The expression \(F_X(T;\theta)-F_X(c;\theta)\) instead describes \(c<X\le T\). For an exact observation, the numerator is \(f_X(x;\theta)\), not \(F_X(x;\theta)\). A CDF is appropriate for an event or interval; a density is appropriate for an exact value in a continuous model.

Another check is whether the censoring point is below the truncation threshold. If \(c<T\), the censored event is a proper subset of the selected range and has conditional probability \(F_X(c;\theta)/F_X(T;\theta)\). If \(c\ge T\), every selected record satisfies \(X\le c\), so the contribution is \(1\). If a proposed observation falls outside the model support or violates the selection rule, the corresponding likelihood is zero rather than a usable contribution.

The key reasoning is simple but consequential: the data describe losses drawn from the original model and then observed only after selection. Thus the likelihood must use the distribution conditional on selection, while censoring determines whether the numerator is a density or an event probability. This separation is the main safeguard against incorrect likelihoods in combined truncation-and-censoring problems.

Key takeaway: With upper truncation at \(T\), divide every record's contribution by \(F_X(T;\theta)\). Use \(f_X(x;\theta)\) in the numerator for an exact loss and \(F_X(c;\theta)\) for a record reported only as \(X\le c<T\).

Check Your Understanding

For each question, treat the stated sample as conditional on selection below its upper truncation threshold. Round numerical probabilities or likelihoods to four decimal places.

A loss is exponential with rate \(\lambda=0.3\) per thousand dollars. The truncation threshold is \(T=5\) thousand dollars, and a sampled record is known only to be at or below \(c=2\) thousand dollars. Find its likelihood contribution. (Answer: \(0.5807\))
A loss is exponential with rate \(\lambda=0.5\) per thousand dollars, truncated above at \(T=3\) thousand dollars. A selected loss is observed exactly at \(x=2\) thousand dollars. Find its conditional density contribution. (Answer: \(0.2368\))
A Pareto Type I loss has minimum \(\theta=2\) thousand dollars and shape \(\alpha=3\). The sample is truncated above at \(T=8\) thousand dollars, and one selected record is known only to satisfy \(X\le4\) thousand dollars. Find its contribution. (Answer: \(0.8889\))
For an exponential loss with rate \(\lambda=0.5\) per thousand dollars, a sample is truncated above at \(T=4\) thousand dollars. Two records are known to be at or below \(c=1\) thousand dollars; two other losses are observed exactly at \(2\) and \(3\) thousand dollars. Write the product of contributions and evaluate it. (Answer: \(\left[(1-e^{-0.5})/(1-e^{-2})\right]^2\left[0.5e^{-1}/(1-e^{-2})\right]\left[0.5e^{-1.5}/(1-e^{-2})\right]\approx0.0057\))
A continuous loss model is sampled only below \(T=2\) thousand dollars. A selected record is reported as \(X\le c\), where \(c=3\) thousand dollars. What is its conditional probability contribution, regardless of the continuous model, assuming \(F_X(2)>0\)? (Answer: \(1\))

What You'll Learn